Magic 8 Ball
Write a
Contents of 8_ball_responses.txt:
Yes, of course !
Without a doubt, yes.
you can count on it.
For sure!
Ask me later.
I'm not sure.
I can't tell you right now.
I'll tell you after my nap.
No way!
I don't think so
Without a doubt, no.
The answer is clearly No.
Want to see the full answer?
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- Refer to page 70 for problems related to process synchronization. Instructions: • • Solve a synchronization problem using semaphores or monitors (e.g., Producer-Consumer, Readers-Writers). Write pseudocode for the solution and explain the critical section management. • Ensure the solution avoids deadlock and starvation. Test with an example scenario. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qo Hazb9tC440AZF/view?usp=sharing]arrow_forward15 points Save ARS Consider the following scenario in which host 10.0.0.1 is communicating with an external SMTP mail server at IP address 128.119.40.186. NAT translation table WAN side addr LAN side addr (c), 5051 (d), 3031 S: (e),5051 SMTP B D (f.(g) 10.0.0.4 server 138.76.29.7 128.119.40.186 (a) is the source IP address at A, and its value. S: (a),3031 D: (b), 25 10.0.0.1 A 10.0.0.2. 1. 138.76.29.7 10.0.0.3arrow_forward6.3A-3. Multiple Access protocols (3). Consider the figure below, which shows the arrival of 6 messages for transmission at different multiple access wireless nodes at times t=0.1, 1.4, 1.8, 3.2, 3.3, 4.1. Each transmission requires exactly one time unit. 1 t=0.0 2 3 45 t=1.0 t-2.0 t-3.0 6 t=4.0 t-5.0 For the CSMA protocol (without collision detection), indicate which packets are successfully transmitted. You should assume that it takes .2 time units for a signal to propagate from one node to each of the other nodes. You can assume that if a packet experiences a collision or senses the channel busy, then that node will not attempt a retransmission of that packet until sometime after t=5. Hint: consider propagation times carefully here. (Note: You can find more examples of problems similar to this here B.] ☐ U ப 5 - 3 1 4 6 2arrow_forward
- Just wanted to know, if you had a scene graph, how do you get multiple components from a specific scene node within a scene graph? Like if I wanted to get a component from wheel from the scene graph, does that require traversing still? Like if a physics component requires a transform component and these two component are part of the same scene node. How does the physics component knows how to get the scene object's transform it is attached to, this being in a scene graph?arrow_forwardHow to develop a C program that receives the message sent by the provided program and displays the name and email included in the message on the screen?Here is the code of the program that sends the message for reference: typedef struct { long tipo; struct { char nome[50]; char email[40]; } dados;} MsgStruct; int main() { int msg_id, status; msg_id = msgget(1000, 0600 | IPC_CREAT); exit_on_error(msg_id, "Creation/Connection"); MsgStruct msg; msg.tipo = 5; strcpy(msg.dados.nome, "Pedro Silva"); strcpy(msg.dados.email, "pedro@sapo.pt"); status = msgsnd(msg_id, &msg, sizeof(msg.dados), 0); exit_on_error(status, "Send"); printf("Message sent!\n");}arrow_forward9. Let L₁=L(ab*aa), L₂=L(a*bba*). Find a regular expression for (L₁ UL2)*L2. 10. Show that the language is not regular. L= {a":n≥1} 11. Show a derivation tree for the string aabbbb with the grammar S→ABλ, A→aB, B→Sb. Give a verbal description of the language generated by this grammar.arrow_forward
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